Web Analytics
Answered

A trailer mechanic pushes a 2500 kg car, home to a speed v, performing a job during the 5000 J. In this process, the car moves 25 m. Neglecting friction between the car and roadway:

AWhat is the final velocity v of the car? (B) What horizontal force exerted on the car?

Verified Answer :

Hope it's visible. All the best :-D

View image Anonymous
shubhamchouhanvrVT

Final velocity  \(V_{f}\)  of the car will be  2 meters per second

Horizontal force exerted on the car will be = 200 newtons

What will be the final speed and horizontal force of the car ?

It is given that mass= 2500 kg

Total energy =5000 j

Car distance = 25 m

So the change in KE \(=\dfrac{1}{2} mv_{f} ^{2} -\dfrac{1}{2} mv_{i} ^{2}\)

                                   \(=\dfrac{1}{2} mv_{f} ^{2}\)       Since \(V_{ i} =0\)

                                   \(=\dfrac{1}{2} \times2500\times v_{f} ^{2}\)

                           \(5000=\dfrac{1}{2} \times2500\times v_{f} ^{2}\)

                                \(V_{f} =2 \dfrac{m}{s}\)

Now the Total change in energy = Work done

                                       \(=F\times d\)

                               \(5000= F\times 25\)

                                   \(F= 200 N\)

Hence

Final velocity  \(V_{f}\)  of the car will be  2 meters per second

Horizontal force exerted on the car will be = 200 newtons

To know more about Work energy follow

https://brainly.com/question/22236101

Related Questions